Total going down timer time if it has chance of increasing I ran into this logical issue and cannot figure out how many tries it should take for time to run out.
I have 100 seconds timer  which goes down every second. Each time it goes down, it has a 1/3rd chance of increasing current time by 1 second.
How many seconds will it have to go through eventually until timer runs out? 
 A: It sounds like you have a $100$ second timer, and every second there is a $\frac 2 3 $ probability that the timer goes down by $1$ and a $\frac 1 3$ probability that it increases by $1$.
It only has to go through 100 seconds to run out, but will take significantly longer than that in the vast majority of cases.  (If there is an incredible streak, it might just tick down every second rather than ever going back up.)
It seems like the expected average time should be $300$ seconds.
  In a given second the expected change in time is $\frac 2 3 \cdot -1 + \frac 1 3 \cdot 1 = \frac {-2} 3 + \frac 1 3 = \frac {-1} 3$.  So in $300$ seconds, we expect to have changed the time by $300 \cdot  \frac {-1} 3 = -100$ seconds.
A: It seems to me like there aren't enough data to solve the problem.  Here is my analysis.
Let $E_n$ be the expected number of ticks to count down to $0$ if the current reading is $n,$ so that we are trying to compute $E_{100}.$  We have $$E_n=1+\frac23 E_{n-1}+\frac13 E_{n+1}\tag{1}$$ and $$E_0=0$$
$(1)$ leads to the three-term recurrence $$E_{n+1}-3E_n+2E_{n-1}=3$$ whose general solution is $$E_n=\lambda_12^n+\lambda_2-3n\tag{2},$$ if I'm not mistaken.  Now substituting in $E_0= 0$ in $(2)$ gives $\lambda_1=-\lambda_2=\lambda$ say, but we need another condition to solve for $\lambda.$ 
For example, if we knew that when the times reads $999$ if it ticks "up" it goes to $0$ and stops, then we could solve for $\lambda,$ I think, though it might get ugly. 
